This may look like a physics problem, but everything physical is explained below and I am looking for a mathematical solution.
An RLC circuit (pictured above) is governed by two equations: $$ -iR=-L \frac{dj}{dt} = \frac{q}{C}+V(t) $$ $$\frac{dq}{dt}=i+j$$
q satisfies the equation: $$\frac{d^2q}{dt^2}+\frac{1}{RC}\frac{dq}{dt}+\frac{1}{LC}q=-\frac{1}{R}\frac{dV}{dt}-\frac{1}{L}V $$ The system is held in a steady state (i.e. $\frac{dq}{dt}=0$ and $V(t)=\frac{Q}{C}$) for negative time. At t=0 the voltage is switched off and $V(t) = 0$ for $t \geq 0$.
Question: how to derive the initial conditions for the system, i.e. $q(0)=Q$ and $\dot{q}(0)=\frac{Q}{RC}$ ?
My attempt: to calculate the charge in the steady state (just before t=0), I can set all derivatives with respect to time to 0. Then I get $V=-\frac{C}{q}$ and I can define $Q=-\frac{C}{V}$. I don't know how to handle the discontinuity at t=0 to obtain $\dot{q}(0)$ though.
I think you need to add the physical condition that $j$ is continuous everywhere. Using that, you get $j(0^+)=0$, and thus $\dot{q}(0^+)=i(0^+)=\dots$.
Or without thinking more about physics, you can argue that a discontinuity of $j$ at $0$ would mean that $\frac{dj}{dt}$ had a $\delta$ peak, which is incorrect because $-L\frac{dj}{dt}$ equals something that cannot have a delta peak even in the approximation that $V$ changes abruptly.